Dividing a Segment into N parts:
Similar Right Triangles
A Mathematical Droodle: What Is This About?

 

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Explanation

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Copyright © 1996-2018 Alexander Bogomolny

Dividing a Segment into N parts:
Similar Right Triangles

Following is a simple algorithm for finding the Nth part of a given segment, N > 1, see [Maor, p. 96].

 

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Let AB be a given segment and C a point on the line AB satisfying AC = N·AB. Draw two circles. One A(B) with center A and radius AB and another O(r), r = AC/2, on AC as a diameter. The circles cross at points E and D, and P is their common projection on line AB. Then AP = AB/N.

Indeed, triangles AEC and APE are right and share an angle at A. They are therefore similar implying a proportion

(1) AP / AE = AE / AC

However, AC = N·AB = N·AE. Thus (1) gives

(2) AP = AE² / AC = AE / N = AB / N.

This is a clear generalization of the standard procedure of dividing a segment into two (N = 2).

References

  1. E. Maor, The Pythagorean Theorem, Princeton University Press, 2007

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Copyright © 1996-2018 Alexander Bogomolny

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